Triangle
This page illustrates the Triangle shape in the Julia package ImagePhantoms.
This page was generated from a single Julia file: 05-triangle.jl.
In any such Julia documentation, you can access the source code using the "Edit on GitHub" link in the top right.
The corresponding notebook can be viewed in nbviewer here: 05-triangle.ipynb, and opened in binder here: 05-triangle.ipynb.
Setup
Packages needed here.
using ImagePhantoms: Triangle, phantom, radon, spectrum
using ImageGeoms: ImageGeom, axesf
using MIRTjim: jim, prompt
using FFTW: fft, fftshift
using Unitful: mm, unit, °
using UnitfulRecipes
using Plots: plot, plot!, scatter!, default; default(markerstrokecolor=:auto)The following line is helpful when running this example.jl file as a script; this way it will prompt user to hit a key after each figure is displayed.
isinteractive() ? jim(:prompt, true) : prompt(:draw);Overview
For completeness, this package includes a triangle shape for constructing 2D digital image phantoms. (One could describe quite complicated phantoms with a triangular mesh.) The basic shape here is an equilateral triangle whose base is [-1/2,1/2] along the x axis, pointing upwards along the y axis. When defining such a Triangle object one can specify its center, widths, angle and value. All of the methods in ImagePhantoms support physical units, so we use such units throughout this example. (Using units is recommended but not required.)
Define a triangle object, using physical units.
width = (2mm, 8mm)
ob = Triangle((4mm, 3mm), width, π/6, 1.0f0)Object2d{Triangle, Float32, 2, Unitful.Quantity{Int64, 𝐋, Unitful.FreeUnits{(mm,), 𝐋, nothing}}, Float64, Float64} (S, D, V, ...)
shape::Triangle Triangle()
center::NTuple{2,Unitful.Quantity{Int64, 𝐋, Unitful.FreeUnits{(mm,), 𝐋, nothing}}} (4 mm, 3 mm)
width::NTuple{2,Unitful.Quantity{Int64, 𝐋, Unitful.FreeUnits{(mm,), 𝐋, nothing}}} (2 mm, 8 mm)
angle::Tuple{Float64} (0.5235987755982988,)
value::Float32 1.0
param::Float64 0.5
Phantom image using phantom
Make a digital image of it using phantom and display it.
dx = 0.02mm
dy = 0.025mm
(M,N) = (2^10,2^10+2)
x = (-M÷2:M÷2-1) * dx
y = (-N÷2:N÷2-1) * dy
img = phantom(x, y, [ob])
jim(x, y, img)Hereafter we use ImageGeoms to simplify the indexing.
ig = ImageGeom(dims=(M,N), deltas=(dx,dy), offsets=(0.5,0.5))
@assert all(axes(ig) .≈ (x,y))
p1 = jim(axes(ig)..., img, "Triangle phantom", xlabel="x", ylabel="y")Spectrum using spectrum
There are two ways to examine the spectrum of this image:
- using the analytical Fourier transform of the triangle via
spectrum - applying the DFT via FFT to the digital image.
Because the shape has units mm, the spectra axes have units cycles/mm.
zscale = 1 / (sqrt(3) / 4 * prod(width)) # normalize spectra by area
spectrum_exact = spectrum(axesf(ig)..., [ob]) * zscale
sp = z -> max(log10(abs(z)/oneunit(abs(z))), -6) # log-scale for display
clim = (-6, 0) # colorbar limit for display
(xlabel, ylabel) = ("ν₁", "ν₂")
p2 = jim(axesf(ig)..., sp.(spectrum_exact), "log10|Spectrum|"; clim, xlabel, ylabel)Sadly fft cannot handle units currently, so this function is a work-around:
function myfft(x)
u = unit(eltype(x))
return fftshift(fft(fftshift(x) / u)) * u
end
spectrum_fft = myfft(img) * dx * dy * zscale
p3 = jim(axesf(ig)..., sp.(spectrum_fft), "log10|DFT|"; clim, xlabel, ylabel)Compare the DFT and analytical spectra to validate the code
@assert maximum(abs, spectrum_exact - spectrum_fft) /
maximum(abs, spectrum_exact) < 2e-2
p4 = jim(axesf(ig)..., abs.(spectrum_fft - spectrum_exact), "Difference"; xlabel, ylabel)
jim(p1, p4, p2, p3)Radon transform using radon
Examine the Radon transform of the triangle using radon, and validate it using the projection-slice theorem aka Fourier-slice theorem.
dr = 0.02mm # radial sample spacing
nr = 2^10 # radial sinogram bins
r = (-nr÷2:nr÷2-1) * dr # radial samples
fr = (-nr÷2:nr÷2-1) / nr / dr # corresponding spectral axis
ϕ = deg2rad.(0:180) # * Unitful.rad # todo round unitful Unitful.°
sino = radon(ob).(r, ϕ') # sample Radon transform of a single shape object
p5 = jim(r, rad2deg.(ϕ), sino; aspect_ratio=:none, title="sinogram", yflip=false, xlabel="r", ylabel="ϕ")The maximum sinogram value is about 7mm, which makes sense for a triangle whose height is 8mm * sqrt(3) / 2 and whose base is 2mm, so the longest side is sqrt(1^2 + (8mm * sqrt(3) / 2)^2) = 7 mm.
The above sampling generated a parallel-beam sinogram, but one could make a fan-beam sinogram by sampling (r, ϕ) appropriately.
Fourier-slice theorem illustration
Pick one particular view angle (55°) and look at its slice and spectra.
ia = argmin(abs.(ϕ .- 55°))
slice = sino[:,ia]
slice_fft = myfft(slice) * dr
angle = round(rad2deg(ϕ[ia]), digits=1)
kx, ky = (fr * cos(ϕ[ia]), fr * sin(ϕ[ia])) # Fourier-slice theorem
slice_ft = spectrum(ob).(kx, ky)
@assert maximum(abs, slice_ft - slice_fft) / maximum(abs, slice_ft) < 2e-4
p3 = plot(r, slice, title="profile at ϕ = $angle", label="")
p4 = plot(title="1D spectra")
scatter!(fr, abs.(slice_fft), label="abs fft", color=:blue)
scatter!(fr, real(slice_fft), label="real fft", color=:green)
scatter!(fr, imag(slice_fft), label="imag fft", color=:red, xlims=(-1,1).*(1.0/mm))
plot!(fr, abs.(slice_ft), label="abs", color=:blue)
plot!(fr, real(slice_ft), label="real", color=:green)
plot!(fr, imag(slice_ft), label="imag", color=:red)
plot(p1, p5, p3, p4)The good agreement between the analytical spectra (solid lines) and the DFT samples (disks) validates that phantom, radon, and spectrum are all self consistent for this Triangle object.
Spectrum
The spectrum of a triangle is not widely available; for a derivation, see this overleaf file.
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